Question

1) Consider the three blocks, A, B, C, resting on the horizontal surface below. The coefficient of frictions between the blocks and the surface are ua= 0.15, Ue = 0.20, and uc= 0.25, respectively. The masses of the blocks are ma= 5.0kg, m3 = 10kg, and mc = 15kg. A horizontal force of 100N is applied to the A block. Draw a relatively accurate FBD for each block. Find the force that block B exerts on block C. В

Answer 1

Answer!) I TBN, * 0 % a, - Normal force b/w A&B Ng Normal force betweem B&C from Newton's and Law fa (N,+ fi) = ma a £ No - (a + f2) = mga

Ng-f3= ma full from 100-(N, * 0.15% mag) = maa 7 100- N-7.35= als) - 100-7.35-N = sa - So. N = 92.65-5 a put in aan & 92.65-5.A - (Ng & fz ) = mo a 99.65 -Sa-Ng - 0.90X 10X9.8= loa - 99.65 - 19.6.-Ag = lo at sa 7 193.05 - Ng = Sao put in on 11 Ngo (0.25 x 15x 9.8) = 15e 36.75 = isa fuam W & N-36.75= 93.05 - Ng 9 Ng = 93.05+ 36.75 2 Ng = 129.8 No = 64.9N) lormal Block B - black C fore which apply on & vic versa
from Newton's second law write their equation of motion and solve for the force which block B applies on Block A